Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class num>
inline void rd(num &x) {
  char c;
  while (isspace(c = getchar()))
    ;
  bool neg = false;
  if (!isdigit(c))
    neg = (c == '-'), x = 0;
  else
    x = c - '0';
  while (isdigit(c = getchar())) x = (x << 3) + (x << 1) + c - '0';
  if (neg) x = -x;
}
const int N = 101234;
struct node {
  vector<int> values;
  node() { values.resize(10, INT_MAX); }
  node(vector<int> x) { values = x; }
  node operator+(node a) {
    node ret;
    int l = 0, r = 0, ct = 0;
    while (ct < 10) {
      ret.values[ct++] = a.values[l] <= values[r] ? a.values[l++] : values[r++];
    }
    return ret;
  }
};
vector<int> v[N];
int ordem[N];
struct segtree {
  vector<node> seg;
  int n, k, a, b;
  segtree() {}
  segtree(int n) : n(n) {
    seg.resize(4 * n + 10, node());
    build(1, 0, n - 1);
  }
  void build(int r, int i, int j) {
    if (i == j) {
      seg[r] = node();
      seg[r].values = v[ordem[i]];
    } else {
      int mid = (i + j) / 2;
      build(r * 2, i, mid);
      build(r * 2 + 1, mid + 1, j);
      seg[r] = seg[r * 2] + seg[r * 2 + 1];
    }
  }
  node query(int r, int i, int j) {
    if (i > b or j < a) return node();
    if (i >= a and j <= b) return seg[r];
    int mid = (i + j) / 2;
    node le = query(r * 2, i, mid);
    node ri = query(r * 2 + 1, mid + 1, j);
    return le + ri;
  }
  node query(int l, int r) {
    a = l;
    b = r;
    return query(1, 0, n - 1);
  }
} st;
int pai[N][20], cost[N], prof[N], sz[N];
int heavy[N];
vector<int> g[N];
int query_lca(int u, int v) {
  if (prof[u] > prof[v]) swap(u, v);
  int diff = prof[v] - prof[u];
  for (int j = 0; j < 20; j++) {
    if (diff & (1 << j)) v = pai[v][j];
  }
  if (u == v) return u;
  for (int j = 19; j >= 0; j--) {
    if (pai[u][j] != pai[v][j]) {
      u = pai[u][j];
      v = pai[v][j];
    }
  }
  return pai[u][0];
}
void dfs(int u, int pu, int p = 0) {
  sz[u] = 1;
  prof[u] = p;
  for (int v : g[u]) {
    if (v == pu) continue;
    dfs(v, u, p + 1);
    pai[v][0] = u;
    sz[u] += sz[v];
    if (sz[heavy[u]] < sz[v]) heavy[u] = v;
  }
}
int cnt_chain, chainId[N];
vector<int> chains[N];
int posChain[N], ent[N], sai[N], dfscnt;
node pref[N];
void hld(int u, int pu, int chain) {
  chains[chain].push_back(u);
  chainId[u] = chain;
  posChain[u] = chains[chain].size() - 1;
  ordem[dfscnt] = u;
  ent[u] = dfscnt++;
  if (posChain[u] != 0) {
    pref[u] = pref[pu] + node(v[u]);
  } else {
    pref[u] = node(v[u]);
  }
  if (heavy[u] != 0) hld(heavy[u], u, chain);
  for (auto v : g[u]) {
    if (v == pu or v == heavy[u]) continue;
    cnt_chain++;
    hld(v, u, cnt_chain);
  }
  sai[u] = dfscnt - 1;
}
node query(int u, int lca, int f = 0) {
  node ans;
  while (true) {
    if (chainId[u] != chainId[lca]) {
      int head = chains[chainId[u]][0];
      ans = ans + pref[u];
      u = pai[head][0];
    } else {
      ans = ans + st.query(ent[lca] + f, ent[u]);
      return ans;
    }
  }
}
int main(void) {
  int n;
  rd(n);
  int m;
  rd(m);
  int q;
  rd(q);
  for (int i = 0; i + 1 < n; i++) {
    int u, v;
    rd(u);
    rd(v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(pai, -1, sizeof pai);
  for (int i = 0; i < m; i++) {
    int x;
    rd(x);
    v[x].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++) v[i].resize(10, INT_MAX);
  dfs(1, 0);
  hld(1, 0, 0);
  st = segtree(n);
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= n; i++) {
      if (pai[i][j - 1] != -1) pai[i][j] = pai[pai[i][j - 1]][j - 1];
    }
  }
  while (q--) {
    int u, v, a;
    rd(u);
    rd(v);
    rd(a);
    int lca = query_lca(u, v);
    node ans = query(u, lca) + query(v, lca, 1);
    vector<int> res;
    for (int i = 0; i < a; i++) {
      if (ans.values[i] == INT_MAX) break;
      res.push_back(ans.values[i]);
    }
    printf("%ld", res.size());
    for (int i = 0; i < res.size(); i++) printf(" %d", res[i]);
    printf("\n");
  }
  return 0;
}